of 1 60
An Approach: Proton-Seconds
Ian Beardsley!
(University of Oregon, Department of Physics, 2022)!
!
Genesis Project California 2022!
⋅
⋅
of 2 60
Introduction………………………………………………..3
The Theory…………………………………………………6
Generating A Table……………………………………….8
The Formulation…………………………………………..10
Solving Problems With Proton-Seconds
Example 1 Radius of Solar System……..…………….10
Solving Problems With Proton-Seconds
Example 2 Radius of Proton……………………………17
Solving Problems With Proton-Seconds
Example 3 Relative Abundances Of
Hydrogen and Helium in The Universe……………….18
Solving Problems With Proton-Seconds
Example 4 Asteroids, Meteors, And Meteorites…….21
Giordano’s Relationship…………………………………24
The Elements………………………………………………30
Conclusion 1……………………………………………….39
The Second…………………………………………………40
Conclusion 2………………………………………………..42
Archaeology of other Star Systems…………………….44
Conclusion 3…………………………………………………59
of 3 60
Introduction
With the method of proton-seconds we can solve a great deal of problems easily. They
define a definition for the radius of the solar system and predict the radius of a proton
within its experimental errors to predicting the relative abundances of hydrogen and
helium in the Universe from which the heavier elements are made in the interior of stars.
It is doing physics with a chemistry methodology.
The method involves:
Step 1
Write down the fundamental equation which is equation 3:
Step 2
Generate a table of values with source code for applying the equation in step 1. Like
using your float values:
float value=0, increment,t=0, p=1.67262E-27, h=6.62607E-34,G=6.67408E-11,
c=299792459,protons[100],r=0.833E-15;!
And, the equation that uses them:
protons[I]=((137*137)/(t*p))*sqrt(h*4*(3.14159)*(r*r)/(G*c));!
And a loop for finding whole number solutions and listing them:
while (n>=101);!
{!
for (int i=0; i<n;i++)!
{!
protons[i]=((137*137)/(t*p))*sqrt(h*4*(3.14159)*(r*r)/(G*c));!
!
int intpart=(int)protons[i];!
float decpart=protons[i]-intpart;!
t=t+increment;!
if (decpart<0.25)!
{ printf("%.4f protons %f seconds %f decpart \n", protons[i], t-increment, decpart);!
}!
1
α
2
m
p
h4πr
2
p
Gc
∫
t
2
t
1
1
t
2
dt = ℕ
of 4 60
Then output the values so you can create a list to apply the fundamental equation,
equation 3:
24.1199 protons 0.250000 seconds 0.119904 decpart !
12.0600 protons 0.500000 seconds 0.059952 decpart !
8.0400 protons 0.750000 seconds 0.039968 decpart !
6.0300 protons 1.000000 seconds 0.029976 decpart !
4.0200 protons 1.500000 seconds 0.019984 decpart !
3.0150 protons 2.000000 seconds 0.014988 decpart !
2.1927 protons 2.750000 seconds 0.192718 decpart !
2.0100 protons 3.000000 seconds 0.009992 decpart !
1.2060 protons 5.000000 seconds 0.205995 decpart !
1.1486 protons 5.250000 seconds 0.148567 decpart !
1.0964 protons 5.500000 seconds 0.096359 decpart !
1.0487 protons 5.750000 seconds 0.048691 decpart !
1.0050 protons 6.000000 seconds 0.004996 decpart !
0.2487 protons 24.250000 seconds 0.248659 decpart !
0.2461 protons 24.500000 seconds 0.246121 decpart !
0.2436 protons 24.750000 seconds 0.243635 decpart !
Step 3
Choose the values in the table of outputted integer values as the limits of integration for
the particular problem:
Or, use non-integer values from a geometric theory
!
!
We have considered and . These come from!
!
!
t
Si
= 0.5secon ds
t
C
= 1second
6
∫
3
2
1
t
2
dt = 6
(
1
2
−
1
3
)
= 0.78
∫
3
2
cos
−1
(x /2)d x =
1
6
(
3π − 6
)
−
π − 4
2 2
= 0.21
2
3
2cos(45
∘
) = 2
2cos(30
∘
) = 3
of 5 60
From 30 degrees to 45 degrees is 15 degrees. The Earth rotates through 360/24 is 15
degrees per hour. In this instance we see we have generated the relative bunches of N2
and O2 in Earth air:
In so far as the second unifies carbon (6 protons) with hydrogen (1 proton) through the
unit of a second as the hydrocarbons the backbones of life we suggest the second is
important as well in terms of the phases of the moon and the earth and that these
determined the calendar system we use. We further suggest there is a connection of this
to the structure found in geometry. We see all of this can be written compactly as:
Where n=3, 4, 5, ,6 ,…
We could evaluate this for n equal to other integers, or even the numbers, but these
produce the special triangles, and geometries we are most interested. Thus we will begin
by pointing out that
3 X 4 X 5 X 6 = 360
The amount of degrees into which we divide a circle and that, as such it approximates
the number of days in a year (1 revolution of the earth around the sun) and thus we see
that
Equation 3.
Represents days as well (The earth moves through about 1 degree a day in its orbit
around the sun) by solving it for n:
Equation 4.
Where,
Which correspond respectively to:
Which are the unit triangle, unit the square, the unit regular pentagon, and the unit
regular hexagon.
0.78N 2 + 0.21O2 = 29.0g/m ol
2cos(π /n) = 1, 2, Φ, 3, . . .
2cos(π /n) = 1, 2, Φ, 3, . . .
days = cos
−1
(y/2)
y = 1, 2,
5 + 1
2
, 3, . . .
n = 3,4,5,6,...
of 6 60
The Theory
In order to present the elements as mathematical structures we need to explain the matter from
which they are made as mathematical constructs. We need a theory for Inertia. I had found
(Beardsley Essays In Cosmic Archaeology. 2021) where I suggested the idea of proton
seconds, that is six proton-seconds, which is carbon the core element of biological life if we
can figure out a reason to divide out the seconds. I found!
Equation 1. !
Where h is Planck’s constant 6.62607E—34 Js, is the radius of a proton 0.833E-15m, G is
the universal constant of gravitation 6.67408E-11 (Nm2)/(kg2), and c is the speed of light
299,792,459 m/s. And is t=1 second. is the Sommerfeld constant (or fine structure
constant) is 1/137. The mass of a proton is .!
The fine structure constant squared is the ratio of the potential energy of an electron in the first
circular orbit to the energy given by the mass of an electron in the Bohr model times the speed
of light squared: !
!
Matter is that which has inertia. This means it resists change in position with a force applied to
it. The more of it, the more it resists a force. We understand this from experience, but what is
matter that it has inertia?!
I would like to answer this by considering matter in one of its simplest manifestations, the
proton, a small sphere with a mass of 1.6726E-27 kg. This is a measure of its inertia.!
I would like to suggest that matter, often a collection of these protons, is the three dimensional
cross-section of a four dimensional hypersphere.!
The way to visualize this is to take space as a two-dimensional plane and the proton as a two
dimensional cross-section of a sphere, which would be a circle.!
In this analogy we are suggesting a proton is a three dimensional bubble embedded in a two
dimensional plane. As such there has to be a normal vector holding the higher dimensional
sphere in a lower dimensional space. Thus if we apply a force to to the cross-section of the
sphere in the plane there should be a force countering it proportional to the normal holding it in
a lower dimensional universe. This counter force would be experienced as inertia. It may even
induce in it an electric field, and we can see how it may do the same equal but opposite for the
electron."
1
t
1
α
2
m
p
h4πr
2
p
Gc
= 6protons
r
p
t
1
α
m
p
= 1.67262E − 27kg
α
2
=
U
e
m
e
c
2
of 7 60
of 8 60
!
!
!
Generating A Table
We make a program that looks for close to whole number solutions so we can create a table of
values for problem solving. I set it at decimal part equal to 0.25. You can choose how may
values for t you want to try, and by what to increment them. Here are the results for
incrementing by 0.25 seconds then 0.05 seconds. Constant to all of this is hydrogen and
carbon. The smaller integer value of seconds gives carbon (6 protons at 1 second) and the
largest integer value of seconds gives hydrogen (1 proton at six seconds) and outside of that
for the other integer values of protons you get are at t>0 and t<1. Equation 1 really has some
interesting properties. Here are two runs of the program( decpart is just me verifying that my
boolean test was working right to sort out whole number solutions):!
By what value would you like to increment?: 0.25!
How many values would you like to calculate for t in equation 1 (no more than 100?): 100!
24.1199 protons 0.250000 seconds 0.119904 decpart !
12.0600 protons 0.500000 seconds 0.059952 decpart !
8.0400 protons 0.750000 seconds 0.039968 decpart !
6.0300 protons 1.000000 seconds 0.029976 decpart !
4.0200 protons 1.500000 seconds 0.019984 decpart !
3.0150 protons 2.000000 seconds 0.014988 decpart !
2.1927 protons 2.750000 seconds 0.192718 decpart !
2.0100 protons 3.000000 seconds 0.009992 decpart !
1.2060 protons 5.000000 seconds 0.205995 decpart !
1.1486 protons 5.250000 seconds 0.148567 decpart !
1.0964 protons 5.500000 seconds 0.096359 decpart !
1.0487 protons 5.750000 seconds 0.048691 decpart !
1.0050 protons 6.000000 seconds 0.004996 decpart !
0.2487 protons 24.250000 seconds 0.248659 decpart !
0.2461 protons 24.500000 seconds 0.246121 decpart !
0.2436 protons 24.750000 seconds 0.243635 decpart !
By what value would you like to increment?: 0.05!
How many values would you like to calculate for t in equation 1 (no more than 100?): 100!
40.1998 protons 0.150000 seconds 0.199837 decpart !
30.1499 protons 0.200000 seconds 0.149879 decpart !
24.1199 protons 0.250000 seconds 0.119904 decpart !
20.0999 protons 0.300000 seconds 0.099918 decpart !
17.2285 protons 0.350000 seconds 0.228500 decpart !
15.0749 protons 0.400000 seconds 0.074938 decpart !
12.0599 protons 0.500000 seconds 0.059950 decpart !
10.0500 protons 0.600000 seconds 0.049958 decpart !
1
α
2
m
p
h4πr
2
p
Gc
= (6protons)(1secon d )
1
α
2
m
p
h4πr
2
p
Gc
= (1proton)(6secon ds)
α
2
=
U
e
m
e
c
2
of 9 60
8.0400 protons 0.750000 seconds 0.039966 decpart !
7.0941 protons 0.850000 seconds 0.094088 decpart !
6.0300 protons 1.000000 seconds 0.029975 decpart !
5.2435 protons 1.150000 seconds 0.243457 decpart !
5.0250 protons 1.200000 seconds 0.024980 decpart !
4.1586 protons 1.450000 seconds 0.158605 decpart !
4.0200 protons 1.500000 seconds 0.019985 decpart !
3.1737 protons 1.899999 seconds 0.173673 decpart !
3.0923 protons 1.949999 seconds 0.092296 decpart !
3.0150 protons 1.999999 seconds 0.014989 decpart !
2.2333 protons 2.699999 seconds 0.233325 decpart !
2.1927 protons 2.749999 seconds 0.192719 decpart !
2.1536 protons 2.799999 seconds 0.153564 decpart !
2.1158 protons 2.849998 seconds 0.115782 decpart !
2.0793 protons 2.899998 seconds 0.079303 decpart !
2.0441 protons 2.949998 seconds 0.044061 decpart !
2.0100 protons 2.999998 seconds 0.009993 decpart !
1.2433 protons 4.850000 seconds 0.243294 decpart !
1.2306 protons 4.900001 seconds 0.230607 decpart !
1.2182 protons 4.950001 seconds 0.218177 decpart !
Here is the code for the program:!
#include <stdio.h>!
#include <math.h>!
int main(int argc, const char * argv[]) {!
!
int n;!
float value=0, increment,t=0, p=1.67262E-27, h=6.62607E-34,G=6.67408E-11,
c=299792459,protons[100],r=0.833E-15;!
!
do!
{!
printf("By what value would you like to increment?: ");!
scanf("%f", &increment);!
printf("How many values would you like to calculate for t in equation 1 (no more than 100?):
");!
scanf("%i", &n);!
}!
while (n>=101);!
{!
for (int i=0; i<n;i++)!
{!
protons[i]=((137*137)/(t*p))*sqrt(h*4*(3.14159)*(r*r)/(G*c));!
!
int intpart=(int)protons[i];!
float decpart=protons[i]-intpart;!
t=t+increment;!
if (decpart<0.25)!
{ printf("%.4f protons %f seconds %f decpart \n", protons[i], t-increment, decpart);!
}}}}!
of 10 60
While I understand that one second is a human invention and can’t be taken as significant, the
equations have meaning in that there are two equations each utilizing a second so they are
connected. But what really makes me wonder is how one can predict carbon, and the other
hydrogen so accurately with the unit of a second. It was a conspiracy on the part of those who
formulated the duration of a second a long time ago to be what it is? The Ancient Egyptians,
The Babylonians, The Julian calendar—who, what, when?!
6 protons gives a little more than a second. This makes a shorter day. We have!
Equation 2. !
H=1.00784 g/mol, carbon = 6 protons!
h=6.62607E-34, r_p=0.833E-15, G=6.67408E-11, c=299,792,459!
The Formulation
We can actually formulate this differently than we have. We had!
!
!
But if t1 is not necessarily 1 second, and t6 is not necessarily six seconds, but rather t1 and t2
are lower and upper limits in an integral, then we have:!
Equation 3. !
This Equation is the generalized equation we can use for solving problems.!
Solving Problems With Proton-Seconds Example 1 Radius of Solar System
While we have!
!
Is a number of protons!
1
6α
2
m
p
h4πr
2
p
Gc
= 1.004996352secon d s
1
t
1
1
α
2
m
p
h4πr
2
p
Gc
= 6protons
1
t
6
1
α
2
m
p
h4πr
2
p
Gc
= 1proton
1
α
2
m
p
h4πr
2
p
Gc
∫
t
2
t
1
1
t
2
dt = ℕ
1
α
2
m
p
h4πr
2
p
Gc
∫
t
2
t
1
1
t
2
dt = ℕ
of 11 60
!
Is proton-seconds. Divide by time we have a number of protons because it is a mass divided
by the mass of a proton. But these masses can be considered to cancel and leave pure
number. We have that!
!
!
Interestingly 78% is the percent of N2 in the atmosphere and 21% is the percent of O2 in the
atmosphere (by volume). These are the primary constituents that make it up. The rest is
primarily argon and CO2. This gives the molar mass of air as a mixture is:!
!
Now interestingly, I have found this connected to the solar system by considering a mixture of
silicon, the primary constituent of the Earth crust, and germanium just below it in the periodic
table, in the same proportions of 78% and 21%. Silicon is also the primary second generation
semiconductor material (what we use today) and germanium is the primary first generation
semiconductor material (what we used first). The silicon is directly below our carbon of one
proton-second, silicon directly below that, and germanium directly below that, in the periodic
table. So we are moving directly down the periodic table in group 14. The density of silicon is
2.33 g/cm3 and that of germanium is 5.323 g/cm3. Let us weight these densities with our 0.21
and 0.78:!
!
Now consider this the starting point for density of a thin disc decreasing linearly from the Sun
to Pluto (49.5AU=7.4E14cm). Thus,…!
!
Thus,…!
!
Or,…!
!
1
α
2
m
p
h4πr
2
p
Gc
6
∫
3
2
1
t
2
dt = 6
(
1
2
−
1
3
)
= 0.78
∫
3
2
cos
−1
(x /2)d x =
1
6
(
3π − 6
)
−
π − 4
2 2
= 0.21
0.78N 2 + 0.21O2 = 29.0g/m ol
0.21Si + 0.78Ge = 4.64124g/cm
3
ρ(r) = ρ
0
(
1 −
r
R
)
M =
∫
2π
0
∫
R
0
ρ
0
(
1 −
r
R
)
rdrdθ
M =
πρ
0
R
2
3
of 12 60
Thus,…!
!
If we add up the masses of the planets in our solar system they are 2.668E30 grams.!
Since!
!
Meaning mixing germanium and silicon in the same proportion that occurs with N2 and O2 in
the atmosphere and with!
!
!
Where!
!
In the first integral. See the following pages to see the computation of the mass of the planets
in the solar system…"
M =
π(4.64124)(7.4E14)
2
3
= 2.661E30
2.661
2.668
(100) = 99.736
6
∫
3
2
1
t
2
dt = 6
(
1
2
−
1
3
)
= 0.78
∫
3
2
cos
−1
(x /2)d x =
1
6
(
3π − 6
)
−
π − 4
2 2
= 0.21
6 =
1
α
2
m
p
h4πr
2
p
Gc
of 13 60
As we can see Jupiter carries the majority of the mass, Saturn a pretty large piece, and
somewhat large Uranus and Neptune. We don’t even list Pluto’s mass. When we consider!
!
!
We are considering and . These come from!
!
!
6
∫
3
2
1
t
2
dt = 6
(
1
2
−
1
3
)
= 0.78
∫
3
2
cos
−1
(x /2)d x =
1
6
(
3π − 6
)
−
π − 4
2 2
= 0.21
2
3
2cos(45
∘
) = 2
2cos(30
∘
) = 3
of 14 60
From 30 degrees to 45 degrees is 15 degrees. The Earth rotates through 360/24 is 15 degrees
per hour. The hour is divided into 60 minutes and minutes into 60 seconds…(Next page)Radius
of the Solar System!
We have said!
!
For a thin disc. Thus we have a definition for the radius of the solar system, :!
Equation 4. !
Where!
Equation 5. !
Equation 6. !
Equation 7. !
Equation 8. !
is the mass of all the planets. We also have that the molar mass of air to the molar mass of
water is approximately the golden ratio. The interesting thing is we determine a definition for
the radius of the solar system and predict the hydrocarbons (backbones of life) all in one fell
swoop.!
In so much as we have a definition for the radius of the solar system!
!
!
M =
πρ
0
R
2
3
R
s
R
s
=
3M
p
π(0.78Ge + 0.21Si )
1
α
2
m
p
h4πr
2
p
Gc
∫
3
2
1
t
2
dt = 0.78
∫
3
2
cos
−1
(x /2)d x = 0.21
air = 0.78N
2
+ 0.21O
2
air
H
2
O
≈ Φ
M
p
R
s
=
3M
p
π(0.78Ge + 0.21Si )
1
α
2
m
p
h4πr
2
p
Gc
∫
3
2
1
t
2
dt = 0.78
of 15 60
!
Through life at its fundamental structure the hydrocarbons!
!
!
Which occurs in air and water that makes life possible!
!
!
We want to consider its ground state, and because Mercury the first planet is so small and
carries little energy we proceed to Venus and describe it in terms of silicon and germanium like
we did with our definition for the radius of the solar system. I find by molar mass we have its
average orbital distance from the sun (0.72 AU) precisely in the following expression:!
Equation 9. !
Where Si=28.09 grams/mole and Ge=72.64 grams/mole. If we rewrite this:!
Equation 10. !
And differentiate!
!
!
And evaluate these for silicon and germanium we have!
∫
3
2
cos
−1
(x /2)d x = 0.21
1
t
1
1
α
2
m
p
h4πr
2
p
Gc
= 6protons
1
t
6
1
α
2
m
p
h4πr
2
p
Gc
= 1proton
air = 0.78N
2
+ 0.21O
2
air
H
2
O
≈ Φ
0.72AU =
1
Ge
2
2SiGe +
Si
3
Ge
1 +
Si
2
Ge
2
r(x, y) =
1
y
2
2x y +
x
3
y
1 +
x
2
y
2
∂r
∂x
=
x
4
+ x
2
y
2
+ 2y
4
y(x
2
+ y
2
)
2
∂r
∂y
=
x(x
4
+ x
2
y
2
+ 2y
4
)
y
2
(x
2
+ y
2
)
2
of 16 60
!
!
Which is:!
!
Thus doing as we did before, we move down group 14 of the periodic table from carbon, to
silicon, to germanium and have:!
C+Si+Ge=12.01+28.09+72.64=112.74g/mol!
We have!
!
Which is Earth Orbit. Plots of our equation for Venus are:!
∂r
∂x
= 0.0022626491
∂r
∂y
= 0.008749699
(0.00226)
2
+ (0.00875)
2
= 0.009037151
AU ⋅ mol
gra m
0.009037
AU ⋅ mol
g
⋅ 112.74
g
mol
= 1.0AU
of 17 60
It may be we want to consider Jupiter the ground state as it carries the majority of the mass of
the solar system. We may have that Venus is the ground state for the terrestrials planets and
Jupiter the ground state for the outer gas giants, as separated by the asteroid belt, the location
where a planet cannot form. This takes us to Mars, which is beyond which it lay.!
Solving Problems With Proton-Seconds Example 2 Radius of Proton
In that we have!
!
And the periodic table of the elements is cyclical with 18 groups and!
!
Then perhaps we are supposed to write!
!
In fact, what if the 3 is supposed to be pi, then!
!
Then we would say that!
k=18/pi=5.7229577951!
The parameter in our constant with the most uncertainty is the radius of a proton . If the 3 is
supposed to be pi, then the radius of a proton becomes:!
!
Which gives!
!
About 95% raw most recent value measured. But, if !
1
α
2
m
p
h4πr
2
p
Gc
∫
t
2
t
1
1
t
2
dt = ℕ
6 =
1
α
2
m
p
h4πr
2
p
Gc
3
α
2
m
p
h4πr
2
p
Gc
∫
t
2
t
1
1
t
2
dt = 18
π
α
2
m
p
h4πr
2
p
Gc
∫
t
2
t
1
1
t
2
dt = 18
r
p
r
p
= kα
2
m
p
Gc
4πh
r
p
= 8.790587E − 16m
of 18 60
!
Is supposed to be 6 and it is supposed to be multiplied by three to give 18 even which we
need for chemistry so we have 18 protons in the last group of the periodic table which is
important because we need argon with 18 protons for predicting valence numbers of elements
in terms of their need to attain noble gas electron configuration. Then we get!
!
This is in very close agreement with the most recent value measured which is!
!
Which is good for chemistry.!
Solving Problems With Proton-Seconds Example 3 Relative Abundances Of Hydrogen
and Helium in The Universe
We consider a Gaussian wave-packet at t=0:!
!
We say that d which in quantum mechanics would be the delocalization length when squared is
. A is the amplitude and we might say it is . We write the wave packet as a
Fourier transform which is:!
!
We use the identity that gives the integral of a quadratic:!
!
Solve the equation!
!
With the initial condition!
!
1
α
2
m
p
h4πr
2
p
Gc
r
p
= 8.288587E − 16m = 0.829f m
r
p
= 0.833 + / − 0.014
ψ (x,0) = Ae
−
x
2
2d
2
Si − C
C
Ga − P
C
ψ (x,0) = Ae
−
x
2
2d
2
=
∫
dp
2π ℏ
ϕ
p
e
i
h
px
∫
∞
−∞
e
−α
2
x+βx
d x =
π
α
e
β
2
4α
iℏ∂
t
ψ (x, t) =
p
2m
ψ (x, t)
ψ (x,0) =
∫
dp ⋅ e
p
2
d
2
2ℏ
2
⋅ e
−
i
h
px
of 19 60
A plane wave is the solution:!
!
Where, !
The wave-packet evolves with time as!
!
Calculate the Gaussian integral of dp!
and !
The solution is:!
where !
We notice here one of the things you can do with equation 3 is integrate from 0.5 sec to 1 sec
and you get one which multiplied by the constant which is six yields six. Now look up 0.5
seconds from the data output from the program and it is silicon, then go to one second and it
is carbon, thus the integral from silicon in time to carbon in time is carbon in protons. Now
consider life as we know it is based on carbon because it has four valence electrons, but it is
not based on silicon, which has four valence electrons as well, because in the presence of
oxygen it readily forms SiO2 (sand or glass) leaving it unavailable to nitrogen, phosphorus, and
hydrogen to make make amino acids the building blocks of life. But silicon can be doped with
phosphorus, boron, gallium, and arsenic to make semiconductors -- transistor technology from
which we can build artificial life (artificial intelligence, AI). We can integrate over many time
ranges to explore millions of more facets to the equation.!
!
!
!
We take our solution for the probability!
!
e
i
ℏ
( px−ϵ( p)t)
ϵ( p) =
p
2
2m
ψ (x, t) =
∫
dp ⋅ e
p
2
d
2
2ℏ
2
⋅ e
−
i
h
( px−
p
2
2m
t)
α =
d
2
2ℏ
2
+
it
2mℏ
β =
i x
ℏ
ψ
2
= exp
[
−
x
2
d
2
⋅
1
1 + t
2
/τ
2
]
τ =
m d
2
ℏ
t
Si
= 0.5secon ds
t
C
= 1second
ψ
2
= exp
[
−
x
2
d
2
⋅
1
1 + t
2
/τ
2
]
of 20 60
And we say it is:!
Equation 11. !
Let’s say x=1 proton which corresponds to 6 seconds. Then the probability we have a proton is
100% or, the absolute value of psi squared equals 1 means one second meaning we have
carbon or, life in other words. But one it does equal:!
!
We should say!
or !
And is almost exactly one if we average it with the following:!
!
The way I am using equation 1 is . We have:!
Equation 12. !
Thus for hydrogen:!
!
=74%.!
For Helium:!
ψ
2
= exp
[
−
Cx
2
(Si − C )
⋅
1
1 + t
2
/τ
2
]
exp
[
3
4
⋅ (1proton)
2
⋅
1
1 + (6seconds)
2
]
= 1.02
d =
Si − C
C
=
4
3
d
2
=
16
9
exp
[
3
4
⋅ (1proton)
2
⋅
1
(6seconds)
2
]
= 0.979382
τ = d
2
ψ
2
= exp −
C
2
x
2
(Si − C )
2
⋅
1
1 +
[
ℏC
2
m(Si − C)
2
]
2
t
2
ψ
2
= exp −
9
16
x
2
⋅
1
1 +
ℏ
2
81
m
2
256
t
2
ψ
2
= (1)exp −
9
16
(1proton)
2
⋅
1
1 +
(0.075)81
(1)256
(6seconds)
2
of 21 60
=26%!
This is interesting because the Universe is about 74% Hydrogen and 24% Helium, the rest of
the elements making up the other 2%. Thus we can say or . We have
multiplied the first by 1 for Hydrogen element 1, and the second by 2 for helium element 2. In a
sense then, the probabilities represent the probability of finding hydrogen and helium in the
Universe. Hydrogen much of the helium were made theoretically in the Big Bang of the big
bang theory. The other elements were synthesized from these by the stars.!
Solving Problems With Proton-Seconds Example 4 Asteroids, Meteors, And Meteorites
Alpha particles (helium nuclei) have a typical kinetic energy of 5MeV and a speed of 15,000,000
m/s. This is 1.6E-12 Joules and 1.5E7 m/s. Note the defining point here is nickel. This
becomes important because most meteors and asteroids are iron-nickel.!
!
This is 1.5E7 meters!
!
The Solar mass is !
!
!
The Earth radius is 6,378km!
!
We have ( )!
!
About 40 pounds. A meteor has to have a size of at least 1 meter in diameter so as not burn up
completely before making it to the ground. That is, the size required for a meteor to become a
meteorite. About 3.3 feet in diameter. By the time it makes it to the ground it is 128-256 grams.
ψ
2
= (2)exp −
9
16
(2proton)
2
⋅
1
1 +
(0.075)81
(4)256
(3seconds)
2
ℏ
2
= 0.075
ℏ = 0.27386
1.5E 7
m
s
⋅ (1second ) = x
1.6E − 12N ⋅ m
1.5E 7m
= 1.0667E − 19N ≈ 1E − 19Ne wtons
M
⊙
= 1.989E30kg ≈ 2E 30kg
1E − 19kg
m
s
2
⋅
1
2E 30kg
= 5E − 50
m
s
2
M =
ar
2
G
M =
(5E − 50m /s
2
)(6,378, 000m)
2
6.67432E − 11
= 3.0E − 26kg
m
p
= 1.67262E − 27kg
M
m
p
= 18.22kg
of 22 60
For stony irons, that is a density of 4 g/cm3 which is 4000 kg/m3 by these figures. The average
density of a so-called asteroidal chondritic meteor is 4200 kg/m3. Thus we have:!
!
We have 128 grams is 0.128kg in the lower limit for a meteorite. If we want to temper the
meteorite with the meteor before it burns up in the lower limit for size by using the geometric
mean we have multiplying both sides by two for second planet Venus:!
!
Using an exact value for the mass of the sun:!
!
!
This is Venus orbit in astronomical units, which is Venus orbit divided by Earth orbit. Venus is
said to be a failed Earth in that she underwent a runaway greenhouse effect. The chondritic
meteoroid is a meteor that came from the asteroid belt. We currently study the asteroid belt to
catalog asteroid orbits to see if any might be on a trajectory to collide with earth, so we can
knock it off such a trajectory and not suffer the fate of the dinosaurs whose extinction was
brought about by such an asteroid landing in the Yucatan. What this amount to is:!
!
Where E_alpha is the energy of an alpha particle, and v_alpha its velocity. R_e is the radius of
the Earth. And,…!
!
Where the c subscript denotes chondritic meteors. Combining equations we have!
!
In work of mine (Essays In Cosmic Archaeology Volume Four, 2021) I explained the relationship
between planetary orbits and the primary semiconductor elements silicon (Si) and germanium
(Ge) with the very accurate result (silicon also being the main component of sand):!
4
3
π(0.5m)
3
⋅ 4200
kg
m
3
= 2,100kg
4,200 ⋅ 0.17357kg = 27kg
M
m
p
= 19.54kg
19.54kg
27kg
= 0.72
E
α
v
α
t
1
⋅
1
M
⊙
R
2
e
G
⋅
1
m
p
= 19.54kg /s
4R
3
c
M
m
⋅ ρ
c
= 27kg
E
α
v
α
1
t
1
R
E
2
M
⊙
1
Gm
p
4R
3
c
ρ
c
M
m
=
19.54
27
= 0.72
of 23 60
!
By molar mass. This gives!
!
If we can get down to this kind of accuracy for Venus orbit using earth size, the mass of the
sun, the lowest energy orbit in the Bohr atom (fine structure constant) through the lower limit of
a meteor to make it to earth surface (become a meteorite) for these small rocks in the asteroid
belt using proton-seconds like this, the proton-seconds might be useful for modeling the solar
system, which has so many complex unmanageable amounts of factors. In essence these
equations outline an approach. Might even be able to use the Schrödinger wave-equation with
these small rocks in the asteroid belt to determine the probability of asteroids being in the earth
region of space at a given time. !
We have the Schrödinger wave equation is
We must solve it for psi and find psi squared.. !
!
Which is !
!
Because psi is a complex function, so we must multiply it with its complex conjugate and take
the absolute value. To find psi for a region around the earth between the distance from the
earth to the moon all the way around the sun; Near earth objects (NEO!s) are classified by such
a region around the earth, would be very difficult to do for the meteors in the asteroid belt a
meter in diameter as used in my calculations considering them as very small quanta
comprising a cloud like electron clouds in an atom, compared to the size of planets, moons,
and asteroids. However there may be another avenue. We have a data estimate. !
The Earth is hit by about 6,100 meteors a year that are large enough to make it to the ground,
which is about 17 everyday according to Gonzalo Tancredi an astronomer at the University of
the Republic in Montevideo, Uruguay. He arrived at this from a database that says that in the
last 95 years 95 people have observed the rare event where a large meteor the size of a
minivan have entered the atmosphere and broke up into small rocks that showered the sky,
many of them landing on rooftops, which is about 8 per year. To arrive at his figure of 6,100 he
1
Ge
2
2SiGe +
Si
3
Ge
1 +
Si
2
Ge
2
= 0.72
E
α
v
α
1
t
1
R
E
2
M
⊙
1
Gm
p
4R
3
c
ρ
c
M
m
=
1
Ge
2
2SiGe +
Si
3
Ge
1 +
Si
2
Ge
2
iℏ
∂ ψ
∂t
=
(
ℏ∇
2
2m
+ V(
r)
)
ψ (
r, t)
ψ (x, y, z)
2
d x d ydz
ψ
2
= [ψ (x, y, z, y)][ψ*(x, y . z, t)]
of 24 60
reasoned that people only inhabit 0.44% of the Earth!s land area which is 0.13% of it surface
area and therefore for every observed impact another 770 that impact the Earth go
unobserved. !
Thus 6,100 meteors a year is a frequency of (6100)/(365 days)(24 hrs)(60 min)(60 sec)=(6100)/
(31536000 seconds per year). This is a frequency of 0.0002 meteors per second. !
!
Is the frequency we can use in plank!s equation%
!
I first estimated our asteroid of radius (0.5 m) to be 2,200 kg. Which gave!
!
Giordano’s Relationship
Warren Giordano writes in his paper The Fine Structure Constant And The Gravitational
Constant: Keys To The Substance Of The Fabric Of Space, March 21, 2019:!
In 1980, the author had compiled a series of notes analyzing Einstein’s geometric to kinematic
equations, along with an observation that multiplying Planck’s constant ‘ ’ by ‘ ’, where ‘ ’
is the Fine Structure Constant, and multiplying by yielded Newton’s gravitational constant
numerically, but neglecting any units.
Let’s do that!
(6.62607E-34Js)(1+1/137)(1E23)=6.6744E-11 Js!
And it works, G is:!
G=6.67408E-11 N(m2/kg2)!
This is like saying gravitational constant masses equal ten to the 23 Planck velocities:!
!
!
Where m1 is a kilogram and v1 is 1 meter per second.!
!
ν = (2E − 4) ⋅ s
−1
1
2
mv
2
= E = hυ
h =
1,100
0.0002
v
2
h
1 + α
α
10
23
Gm
1
= 10
23
h(1 + α)(v
1
)
Gm
1
= 10
23
h(1 + α)(v
1
)
6E 23 = N
A
of 25 60
But, if we are working with kilograms we say!
!
Is Avogadro’s number, so we have!
!
Which is to say!
!
But H=0.001 kg/mol not 1.00 g/mol and carbon is C=0/0.012 kg/mol not 12.01 g/mol!
Thus!
!
Thus if H is the molar mass of hydrogen (1 g/mol=0.001kg/mol) then there are 6E26 atoms in a
kilogram of it.!
!
Thus saying!
!
Remembering we now say in this system of units if we are working in
kg.!
If we say!
!
And our equations becomes!
Equation 13. !
Equation 14. !
N
A
= 6E 20
Gm
1
=
N
A
6
h(1 + α)(v
1
)
6 ⋅ seconds
kilogra ms
meter
= h
N
A
G
(1 + α)
6 ⋅ seconds
kilogra ms
meter
= h
1000N
A
G
(1 + α)
6 ⋅ seconds
kilogra ms
meter
=
1
1000
h
N
A
G
(1 + α)
6 ⋅ seconds
kilogra ms
meter
= 1000h
N
A
G
(1 + α)
N
A
= 6E 20atom s /m ol
N
A
H
= 6E 23
6 ⋅ seconds
kilogra ms
2
meter
= h
N
A
H
(1 + α)
G
6 ⋅ kilogra ms
2
seconds
meter
= h
N
A
E
(1 + α)
G
of 26 60
These are the variables:!
!
Where is Avogadro’s number and E is the molar mass of an element. Velocity on the lefts
inverted (seconds per meter) is velocity as a frequency. We have the following Schematic (next
page)…"
N
A
E
N
A
of 27 60
!
Equation 2 gives 6 kg squared seconds per meter when evaluated for hydrogen and 1 kg
squared seconds per meter evaluated for carbon. This relationship between 6 and 1 is the
same I got for these same two elements as well in my paper Proton Seconds and Quanta
which was !
Equation 15. !
Equation 16. !
1
α
2
m
p
h4πr
2
p
Gc
= (6protons)(1secon d )
1
α
2
m
p
h4πr
2
p
Gc
= (1proton)(6secon ds)
of 28 60
!
Combining equation 14 with equation 16 (Giordano’s reationship my proton-seconds):!
Equation 17. !
Where is a number of protons. and are meters 1 and kilograms 1 respectively. The
subscripts can increment by integer values for different values.!
Taking equation 14!
!
And Writing it!
Equation 18. !
And looking back at equation 3:!
!
We see they are the same because!
Equation 19. !
But on the left side are and on the right. Which says square
kilograms per meter are protons. This is the Divergence Theorem (Next page)…"
α
2
=
U
e
m
e
c
2
1
(1 + α)
1
α
2
r
p
m
p
G 4π
hc
=
x
1
m
2
1
N
A
E
ℤ
ℤ
x
1
m
1
6 ⋅ kilogra ms
2
seconds
meter
= h
N
A
E
(1 + α)
G
h
(1 + α)
G
N
A
∫
0.5
1.0
1
E
2
dE = 6
1
α
2
m
p
h4πr
2
p
Gc
∫
t
2
t
1
1
t
2
dt = ℕ
h
(1 + α)
G
N
A
=
1
α
2
m
p
h4πr
2
p
Gc
kg
2
m
⋅ s
proton − secon d s
of 29 60
!
This is to say, Since we can write!
h{(1+𝛼)/G}Nª∫{dE/E²}=ℕ!
And, we had!
1/(𝜶²m_p)√{4𝜋h)(𝐫p²)/Gc}∫{dt/t²}=ℕ!
Where!
h{(1+𝛼)/G}Nª=1/(𝜶²m𝑝)√{4𝜋h(r𝑝²)/Gc}!
Then…!
∭dxdydz=(1-0)(1-0)(6-0)=6(kg²)m!
Which is the divergence theorem…!
∭(𝛻∙𝐮)dV=∯𝒖⋅ds"
of 30 60
The Elements
Stokes Theorem, Divergence Theorem, And Proton-SecondsThe theorems from vector
calculus progress from electric and magnetic fields, to protons, to the elements in
structure.
Stokes Theorem……………………………..31
Conclusion 1………………………………….39!
of 31 60
Stokes Theorem
In earlier work I had found:!
Equation 20. !
Where!
Equation 21. !
We noticed we can make a 3 by 3 matrix of it, which lends itself to to the curl of a vector field
by including biological elements carbon C (above Si):!
!
=!
=!
!
Which resulted in Stokes theorem (Beardsley, Essays In Cosmic Archaeology Volume 3):!
Equation 22. !
π
∫
Ge
Si
∫
Ge
Si
(∇ ×
u ) ⋅ d
a =
5
∏
i=1
x
i
5
∏
i=1
x
i
= Si ⋅ Ge ⋅ C ⋅ Ag ⋅ Au
i
j
k
∂
∂x
∂
∂y
∂
∂z
(−C ⋅ P)y (Si ⋅ G a)z (Ge ⋅ As)y
(Ge ⋅ As − Si ⋅ Ga)
i + (C ⋅ P)
k
[
(72.64)(74.92) − (28.09)(69.72)
]
i +
[
(12.01)(30.97)
]
k
5
∫
Ge
Si
∫
Ge
Si
∇ ×
u ⋅ d
a = ex p
(
1
Ge − Si
∫
Ge
Si
ln(x)d x
)
of 32 60
Where!
!
!
!
=(28.085)(72.64)(12.085)(107.8682)(196.9657)=!
!
Showing The Calculation using the most accurate data possible…!
Ge=72.64!
As=74.9216!
Si=28.085!
Ga=69.723!
C=12.011!
P=30.97376200!
=!
=!
!
!
!
!
!
∇ ×
u = (Ge ⋅ As − Si ⋅ Ga)
i + (C ⋅ P)
k
d
a =
(
zd yd z
i + yd yd z
k
)
u = (−C ⋅ P)y
i + (Si ⋅ Ge)z
j + (Ga ⋅ As)y
k
523,818,646.5
g
5
mol
5
(Ge ⋅ As − Si ⋅ Ga)
i + (C ⋅ P)
k
[
(72.64)(74.9216) − (28.085)(69.723)
]
i +
[
(12.011)(30.97376200)
]
k
3,484.134569
(
g
mol
)
2
i + 372.025855
(
g
mol
)
2
k
∫
Ge
Si
∫
Ge
Si
(
3,484.134569
(
g
mol
)
2
i + 372.025855
(
g
mol
)
2
k
)
⋅
(
zd yd z
i + yd yd z
k
)
∫
Ge
Si
∫
Ge
Si
(
3,484.134569
(
g
mol
)
2
⋅ zd zd y + 372.025855
(
g
mol
)
2
⋅ ydzd y
)
∫
Ge
Si
3,484.134569
(
(72.64 − 28.085)
2
2
)
dy +
∫
Ge
Si
372.025855y ⋅ (72.64 − 28.085)d y
3458261.42924
(
g
mol
)
4
(72.64 − 28.085) + 16575.6119695
(
g
mol
)
3
(
(72.64 − 28.085)
2
2
)
of 33 60
=154,082,837.980+16,452,521.6822=!
!
=!
(28.085)(72.64)(12.085)(107.8682)(196.9657)=!
!
Where we have substituted carbon C=12.01 for copper Cu. We use Cu, Ag, Au because they
are the middle column of our electronics matrix, they are the finest conductors used for
electrical wire. We use C, Si, Ge because they are the middle column of our AI Biomatrix. Si
and Ge are the primary semiconductor elements used in transistor technology (Artificial
Intelligence) and C is the core element of biological life. We have!
!
!
Perimeter/Diameter of regular hexagon = 3.00!
!
The same value as our 3.0716 if taken at two places after the decimal.!
The Dynamic Function
The primary elements of artificial intelligence (AI) used to make diodes and transistors, silicon (Si) and
germanium (Ge) doped with boron (B) and phosphorus (P) or gallium (Ga) and arsenic (As) have an
asymmetry due to boron. Silicon and germanium are in group 14 like carbon (C) and as such have 4
valence electrons. Thus to have positive type silicon and germanium, they need doping agents from group
13 (three valence electrons) like boron and gallium, and to have negative type silicon and germanium they
need doping agents from group 15 like phosphorus and arsenic. But where gallium and arsenic are in the
same period as germanium, boron is in a different period than silicon (period 2) while phosphorus is not
(period 3). Thus aluminum (Al) is in boron’s place. This results in an interesting equation.
Equation 23.
The differential across germanium crossed with silicon plus the differential across silicon crossed with
germanium normalized by the product between silicon and germanium is equal to the boron divided by
170,535,359.662
(
g
mol
)
5
5
∏
i=1
x
i
= Si ⋅ Ge ⋅ C ⋅ Ag ⋅ Au
523,818,646.5
g
5
mol
5
523,818,646.5
170,535,359.662
= 3.0716
π = 3.141...
3.141 + 3.00
2
= 3.0705
Si(As − G a) + G e(P − Al )
SiG e
=
2B
Ge + Si
of 34 60
the average between the germanium and the silicon. The equation has nearly 100% accuracy (note: using
an older value for Ge here, it is now 72.64 but that makes the equation have a higher accuracy):
Due to an asymmetry in the periodic table of the elements due to boron we have the harmonic
mean between the semiconductor elements (by molar mass):
Equation 24.
This is Stokes Theorem if we approximate the harmonic mean with the arithmetic mean:
We can make this into two integrals:
Equation 25.
Equation 26.
If in the equation (The accurate harmonic mean form):
Equation 27.
We make the approximation
Equation 28.
Then the Stokes form of the equation becomes
28.09(74.92 − 69.72) + 72.61(30.97 − 26.98)
(28.09)(72.61)
=
2(10.81)
(72.61 + 28.09)
0.213658912 = 0.21469712
0.213658912
0.21469712
= 0.995
Si
B
(As − G a) +
Ge
B
(P − Al ) =
2SiG e
Si + G e
∫
S
(∇ × u ) ⋅ d S =
∮
C
u ⋅ d r
∫
1
0
∫
1
0
[
Si
B
(As − Ga) +
Ge
B
(P − Al )
]
d x d y ≈
1
Ge − Si
∫
Ge
Si
x d x
∫
1
0
∫
1
0
Si
B
(As − Ga)d yd z ≈
1
3
1
(Ge − Si )
∫
Ge
Si
x d x
∫
1
0
∫
1
0
Ge
B
(P − Al )d x dz ≈
2
3
1
(Ge − Si )
∫
Ge
Si
yd y
Si
B
(As − Ga) +
Ge
B
(P − Al ) =
Ge − Si
Ge
Si
dx
x
2SiGe
Si + Ge
≈ Ge − Si
of 35 60
Equation 29.
Thus we see for this approximation there are two integrals as well:
Equation 30.
Equation 31.
For which the respective paths are
One of the double integrals on the left is evaluated in moles per grams, the other grams per mole
(0 to 1 moles per gram and 0 to 1 grams per mole).
By making the approximation
In
We have
Equation 32.
is the differential across Si, is the differential across
Ge and is the vertical differential.!
Which is Ampere’s Circuit Law
∫
1
0
∫
1
0
[
Si
B
(As − Ga) +
Ge
B
(P − Al )
]
d yd z =
∫
Ge
Si
d x
∫
1
0
∫
1
0
Si
B
(As − Ga)d yd z =
1
3
∫
Ge
Si
dz
∫
1
0
∫
1
0
Ge
B
(P − Al )d yd z =
2
3
∫
Ge
Si
dz
y
1
=
1
3
B
SiG a
ln(z)
y
2
=
2
3
B
Si Al
ln(z)
2SiGe
Si + Ge
≈ Ge − Si
Si(As − G a)
B
+
Ge(P − Al )
B
=
2SiGe
Si + Ge
Si
ΔGe
ΔS
+ Ge
ΔSi
ΔS
= B
ΔSi = P − Al
ΔGe = As − Ga
ΔS = Ge − Si
∇ ×
B = μ
0
J + μ
0
ϵ
∂
E
∂t
of 36 60
We see if written
Which is interesting because it is semiconductor elements by molar mass which are used to make
circuits.
We say (Phi) is given by
and
And
=1.618
=0.618
(phi) the golden ratio conjugate. We also find
Equation 13.
Thus since
And we have
Equation 14.
Si
ΔGe
ΔS
= B − Ge
ΔSi
ΔS
Φ
a = b + c
a
b
=
b
c
Φ = a /b
ϕ = b /a
ϕ
(ϕ)ΔGe + (Φ)ΔSi = B
∇ ×
B = μ
J + μϵ
0
∂
E
∂t
Si
ΔGe
ΔS
= B − Ge
ΔSi
ΔS
ΔGe =
ΔS
Si
B −
Ge
Si
ΔSi
(
∇
2
−
1
c
2
∂
2
∂t
)
E = 0
(
∇
2
−
1
c
2
∂
2
∂t
)
B = 0
of 37 60
We see and are both and c is in the Si (silicon) field wave, but for E and B fields c is the speed of
light.
To find the Si wave our differentials are
It is amazing how accurately we can fit these differentials with and exponential equation for the upward
increase. The equation is
This is the halfwave:
c =
1
ϵ
0
μ
≈ ϕ
μ
ϵ
0
Φ
ϕ
ϵ
0
= 8.854E − 12F ⋅ m
−1
μ = 1.256E − 6H /m
Ge
Si
= μϵ
0
ΔS
Si
= μ
(
∇
2
−
1
ϕ
2
∂
2
∂x
)
Si = 0
(
∇
2
−
1
ϕ
2
∂
2
∂x
)
Ge = 0
ΔC = N − B = 14.01 − 10.81 = 3.2
Δ Si = P − Al = 30.97 − 26.98 = 3.99
ΔG e = As − G a = 74.92 − 69.72 = 5.2
Δ Sn = Bi − In = 121.75 − 114.82 = 6.93
ΔPb = Bi − T l = 208.98 − 204.38 = 4.6
y(x) = e
0.4x
+ 1.7
y(x) = e
2
5
x
+
17
10
of 38 60
Equation 33.
Interestingly, the 0.4 is boron (B) over aluminum (Al) the very two elements that lead us to looking for a
wave equation because boron was the out of place element in the AI periodic table that lead to us using
aluminum as its dummy representative in the Si differential and that itself divided into the left hand terms
to give us the harmonic mean between the central AI elements semiconductor materials Si and Ge. The
Ag and Cu are the central malleable, ductile, and conductive metals used in making electrical wires to
carry a current in AI circuitry.
y(x) = e
0.4x
+ 1.7
y(x) = e
2
5
x
+
17
10
y(x) = e
B
Al
x
+
Ag
Cu
B
Al
=
10.81
26.98
= 0.400667
Ag
Cu
=
107.87
63.55
= 1.6974 ≈ 1.7
of 39 60
Conclusion 1
Since!
!
!
Is Stokes Theorem, and!
!
!
Is The Divergence Theorem, and we have Maxwell’s equations!
Then the theorems of vector calculus progress from electric and magnetic fields to
Protons then to the elements in structure.
π
∫
Ge
Si
∫
Ge
Si
(∇ ×
u ) ⋅ d
a =
5
∏
i=1
x
i
Si
B
(As − Ga) +
Ge
B
(P − Al ) =
Ge − Si
Ge
Si
dx
x
1
α
2
m
p
h4πr
2
p
Gc
∫
t
2
t
1
1
t
2
dt = ℕ
h
(1 + α)
G
N
A
∫
0.5
1.0
1
E
2
dE = 6
∮
E ⋅ d
a =
Q
ϵ
0
∮
E ⋅ d
S = −
d
dt
∫
B ⋅ d
a
∮
B ⋅ d
a = 0
c
2
∮
B ⋅ d
S =
d
dt
∮
E ⋅ d
a +
1
ϵ
0
∫
S
J ⋅ d
a
of 40 60
The Second
That,…
Is the unit of a second so closely, and in terms solves so much in a diverse number of
fields, we might consider the second…….
….Actually the second might have physical meaning beyond what is here. Not only is the
second related to the earth orbital period but, to other things; By dividing the day into 24 hours,
the hour into 60 minutes, and the minute into 60 seconds, the second is 1/86400 of day. By
doing this we have a twelve-hour daytime at spring and fall equinox on the equator, 12 being
the most divisible number for its size (smallest abundant number). That is to say that twelve is
evenly divisible by 1,2,3,4,6 which precede it and 1+2+3+4+6=16 is greater than twelve. As
such there is about one moon per 30 days and about 12 moons per year (per each orbit) giving
us a twelve-month calendar. This is all further convenient in that the moon and earth are in very
close to circular orbits and the circle is evenly divisible by 30, 45, 60, and 120 if we divide the
circle into 360 degrees which are special angles very useful to the workings of physics and
geometry. Further, the 360 degrees of a circle are about the 365 days of a year (period of one
earth orbit) so as such the earth moves through about a degree a day in its journey around the
sun. Thus, through these observations down through the ages since ancient times we have
constructed the duration of a second wisely enough to make a lot work together. Now we see 6
protons, which is carbon the core element of biological life on the planet where all of this came
together is deeply connected with the second that defines it all. With this idea of proton-
seconds describing hydrogen and carbon the basis of life the hydrocarbons in a cycle of 6 with
respect to one another, the motion of the earth around the sun and moon around the earth, and
the basis of geometry the 360 degree circle, equation 1 connects them with the universal
constant of gravitation, the speed of light, the fine structure constant and Planck’s constant
that characterize the physics of the atom. I really wonder if other star systems are connected
so well to their planets with their star as is the Earth with the Sun. We really can’t resolve
planets around other stars because the stars are too far and so bright compared to their
orbiting planets and the planets have to be large enough to be inferred by the motion they
induce in the star they orbit. It is hard to do so with earth sized planets that might harbor life.!
It is a mind blower that the constant on page 17 predicts the radius of a proton within the
experimental errors which are within 14 thousandths of a femtometer using a time value of
exactly 1 second because the second originated over ancient times as a means of parsing the
phases of the earth and the moon. This is an archaeological problem and the reason for it might
be deeply profound.
Under the hood a computer is a series of switches that are either on or off. Off is zero and on
is one. With zeros and ones you can create any number:!
0 = 0!
1 =1!
10 = 2!
11 = 3!
100 = 4!
101 = 5!
1
6α
2
m
p
h4πr
2
p
Gc
= 1.004996352secon d s
of 41 60
110 = 6!
111 =7!
And so on,…!
With numbers in binary you can represent any character on your keyboard, numbers or letters.
The numbers assigned to characters, from which your source code is built, are defined in the
ASCII tables, the standardized tables of the language of computer scientists.
The oldest formulated calendar that is known of was written out by the Sumerians and
reconciled the orbit of the moon to the orbit of the earth and was done in the way we still do it
today by inheritance from them by making 12 moons (solar months) per one sun (one year or
completion of the earth around the Sun). The 24 hour day came to us from the Babylonians
and minutes and seconds did not come to us until much later by the Ancient Greeks. The
further division of hours into 60 minutes and minutes into 60 seconds was initiated by Claudius
Ptolemy in his Almagest who divided the latitudes of the earth by 60 to make minutes, and
those by 60 to make seconds. But ultimately dividing things by 60 ultimately comes from the
the Babylonians who inherited it from the Sumerian counting system which was sexagesimal
(base 60) perhaps because of its convenience for dividing fractions because it is the smallest
number divisible by 1, 2, 3. 4, 5, 6, 12, 10, 15, 20, 30 evenly. The second is now more precisely
defined by 9,192,631,770 energy transitions in a cesium atom (1967) which began atomic time
keeping and Coordinated Universal Time (UTC).!
Often used by peoples is what we use today; decimal or base ten because we have ten fingers
to count on, or duodecimal (base 12) because we have 12 joints on our fingers to count, three
for each finger. Thus it all comes from the hand. The twelve hand joints on the hand and 10
fingers taking our two hands together add to make 24 like 24 hours in a day if we bring in the
two joints of the thumbs. As well two hands is the 2 of binary. I would like to suggest we
include a base 14 system by adding the two joints on the thumb to consider archaeologically.!
On Thursday, January 23, 2014 I put forward the following question:!
Are our common usage of variables connected to Nature and the Universe?* I began to look at
this with the following entry: !
We look at (x, y, z,) as they represent the three axis in rectangular coordinates.*We look at* (i, j,
k) as as they are the representations for the unit vectors, and they correspond respectively to
(x, y, z).* We look at*(a, b, c) which has the same kind of correspondence with (x, y, z) as (i, j, k).
All three sets, then, line up with one another and are at the basis of math and physics:!
(a, b, c) corresponds to (x, y, z) corresponds to <i, j, k>.!
(a, b, c) in ASCII computer code is (97, 98, 99) the first three numbers before a hundred and
100 is totality (100%).!
(i, j, k) in numeric are is (9, 10, 11) the first three numbers before twelve and 12 is totality in the
sense that 12 is the most abundant number for its size!
(divisible by 1,2, 3, 4, 6 = 16 is larger than 12).!
(x, y, z) in ASCII computer code is (120, 121, 122) the first three numbers before 123 and 123 is
the number with the digits 1, 2, 3 which are the numeric numbers for the!
(a, b, c) that we started with. These sets of variables, unit vectors, and numbers came before
the advent of the development of the computer and its ascii codes.!
of 42 60
Our system developed historically in binary. Zero is a bit and one is bit. The characters on the
keyboard are described by a byte which is eight bits. This makes possible available
codes. Characters 0-31 are the unprintable control codes used to control peripherals.
Characters 32-127 are printable characters. Capital A to Capital Z are codes 65-90 because
codes 32-64 are taken up for exclamation, comma, period, space and so on,…This puts lower
case a to lower case z at codes 97 to 122.!
What we are saying is if our constant of equation 1 predicts the radius of the proton and one
second is the reason why, then since one second is entwined in a sexagesimal (base 60)
calendar which comes from the orbital periods of the earth and the moon, then the formation of
the planets under gravity on the protoplanetary disc from which they formed is tied up in ,
because it is !
!
This makes sense because while the gravitational interaction between the moon and the earth
is simply a two-body problem.!
!
The interaction between three bodies like for earth and those in its neighborhood Venus and
Mars is a three body problem. There is a general solution of the two body-problem but not the
n-body problem. However the three body problem has a general solution for a particular
scenario. It is not easy because you have!
, , !
Which all needs to be solved simultaneously.!
Conclusion 2
We characterize our mathematics with the following basis…!
Where n=3, 4, 5, ,6 ,…!
We could evaluate this for n equal to other integers, or even the numbers, but these produce
the special triangles, and geometries we are most interested. Thus we will begin by pointing
out that!
3 X 4 X 5 X 6 = 360!
Which, represents days as well (The earth moves through about 1 degree a day in its orbit
around the sun) by solving it for n:!
2
8
= 256
3
2cos(30
∘
) = 3
F = − G
Mm
r
2
F
12
= − G
M
1
M
2
r
2
12
F
13
= − G
M
1
M
3
r
2
13
F
23
= − G
M
2
M
3
r
2
23
2cos(π /n) = 1, 2, Φ, 3, . . .
of 43 60
Where,!
This explains the 360 day year, the 12 hour day, but leaves us wondering about further dividing
things by 60, the hour into 60 minutes, the minute into 60 seconds… !
But ultimately dividing things by 60 ultimately comes from the the Babylonians who inherited it
from the Sumerian counting system which was sexagesimal (base 60) perhaps because of its
convenience for dividing fractions because it is the smallest number divisible by 1, 2, 3. 4, 5, 6,
12, 10, 15, 20, 30 evenly. This inspired me to write a paper titled Archeology of Other Star
Systems…
days = cos
−1
(y/2)
y = 1, 2,
5 + 1
2
, 3, . . .
of 44 60
Archaeology Of Other Star Systems!
of 45 60
Abstract:
A lot comes together for the Earth orbiting the sun to let us know there is a mystery before us if
we look at the archaeology of Earth’s astronomy. Thus, does a lot come together considering
the archaeology of other star systems as well that can indicate to its intelligent life that they are
part of something larger than themselves, as well."
of 46 60
Star System Archaeology
Kepler’s Law of planetary motion is !
Equation 1. !
For the Sun with T the orbital period of the planet and its distance from the Sun in
astronomical units, for circular orbits. For other stars we have to include a constant k involving
the masses of the bodies:!
Equation 2. !
If the mass of the body orbiting the Star m is is small compared to the mass of the star it is
orbiting we have!
Equation 3. !
The Earth is the third planet, is in the habitable zone, and its distance from the Sun defines 1
AU. Thus we ask: What is k for other star systems? For stars on the main sequence their
luminosity is proportional to their masses raised to the power of 3.5 as an estimate. We have:!
Equation 4: !
Where L is in solar luminosities and M is in solar masses:!
!
!
Further if we say since the Earth is in the right zone to be habitable ( ) then if a star is 100
times brighter than the Sun by the inverse square law its habitable zone is is
10 times further from the star it orbits than the Earth is from the Sun. We have!
Equation 5: !
Combining equations 4 and 6:!
Equation 6: !
T
2
= a
3
a
k = G
M + m
4π
2
≈ G
M
4π
2
a
3
=
GM
4π
2
T
2
L =
(
M
M
⊙
)
3.5
L
⊙
= 3.9E 26J/s
M
⊙
= 1.98847E30kg
ℍ
100 = 10AU
ℍ =
(
L
L
⊙
)
ℍ =
(
M
M
⊙
)
3.5
of 47 60
For another star system we can write equation 3!
Equation 7. !
Where n is the number of solar masses of the star. Combining this with equation 6:!
Equation 8. !
Thus if the star is 100 times more luminous than the sun!
!
Or,…!
!
Yielding from equation 6!
!
And!
Equation 9. !
Which is 61 years for 3.72759. This result is (61 yr)(365.25 days)=22,280.25 days for the orbital
period. 10 AU from this star is about the distance Saturn is from the Sun (9.54AU), with year
equal to 10,759 days. The orbital period is longer because it orbits a bit further from its star.!
If in the planet that has life orbiting a star has an indication to its intelligence that there is a
mystery before it, just as we do on the Earth in that we have a moon that perfectly eclipses the
Sun as seen from the Earth, then…!
This is because:!
!
!
Which are approximately equal. As well we can look at it as:!
T
2
a
3
= nk
T
2
= nk
(
M
M
⊙
)
21
4
ln(100) = 3.5ln(M /M
⊙
)
M
M
⊙
= 3.72759
ℍ = 3.72759
3.5
= 10AU
T =
(
M
M
⊙
)
5/2
(lun ar − orbit)
(ear th − orbit)
=
384,400k m
149,592,870k m
= 0.00257
(lun ar − radius)
solar − radiu s
=
1,738.1
696,00
= 0.0025
of 48 60
!
!
Which are about the same as well. The interesting thing is that since our ratios are around
0.0025 and 0.0045, then…!
!
I say this is interesting because this the ratio of the precious metal gold (Au) to that of silver
(Ag) by molar mass these elements being used for religious and ceremonial jewelry:!
!
We have:!
!
!
Taking the average between these:!
Equation 10. !
Where, is the lunar orbit, is the solar radius, is the earth orbit, and is the radius of
the moon. What this means is that the moon perfectly eclipses the sun because the solar
radius is 1.8 times greater than the lunar orbital radius. And interestingly gold is yellow, silver is
silver and the sun is gold, and the moon is silver, the moon perfectly eclipses the sun allowing
us to study its outer atmosphere.!
The Calendar of a People’s Star System
By dividing the day into 24 hours, the hour into 60 minutes, and the minute into 60 seconds,
the second is 1/86400 of day. By doing this we have a twelve-hour daytime at spring and fall
equinox on the equator, 12 being the most divisible number for its size (smallest abundant
number). That is to say that twelve is evenly divisible by 1,2,3,4,6 which precede it and
1+2+3+4+6=16 is greater than twelve. As such there is about one moon per 30 days and about
12 moons per year (per each orbit) giving us a twelve-month calendar. This is all further
convenient in that the moon and earth are in very close to circular orbits and the circle is evenly
divisible by 30, 45, 60, and 120 if we divide the circle into 360 degrees which are special angles
very useful to the workings of physics and geometry. Further, the 360 degrees of a circle are
(lun ar − radius)
(lun ar − orbit)
=
(1,738.1)
(384,400)
= 0.00452
solar − radiu s
ear th − orbit
=
696,000
149,597,870
= 0.00465
0.0045
0.0025
=
9
5
= 1.8
Au
Ag
=
196.97
107.87
= 1.8
(lun ar − radius)(ear th − orbit)
(lun ar − orbit)
2
= 1.759577590
(solar − radiu s)
2
(ear th − orbit)(lunar − radius
= 1.863
1
2
⋅
(
r
2
m
⋅ R
2
⊙
+ r
2
e
R
2
m
r
2
m
⋅ r
e
⋅ R
m
)
=
Au
Ag
r
m
R
⊙
r
e
R
m
of 49 60
about the 365 days of a year (period of one earth orbit) so as such the earth moves through
about a degree a day in its journey around the sun. Thus, through these observations down
through the ages since ancient times we have constructed the duration of a second wisely
enough to make a lot work together.!
Essentially as the moon orbits the earth it makes 12 revolutions for each revolution of the Earth
around the sun which is 365.25 days. That is to say!
!
!
These are frequencies of!
!
!
In radians per day these are:!
!
!
Thus the equations of their phases are:!
!
!
Where t is in days and 2.55605E-3=!
(radius lunar orbit)(radius earth orbit)=384,400km/(149,597,876km).!
We can say the frequency of the moon is 0.21/0.0172=12.21 times greater than that of the
earth. Thus we have the following plots of lunar phases to earth phases:!
T
e
= 365.25days
T
m
= 29.53059days
f
e
= 0.002737851s
−1
f
m
= 0.033863191d ays
−1
ω
e
= 0.0172
ω
m
= 0.21
y
e
= cos(ω
e
t)
y
m
= (2.57E − 3)cos(ω
m
t)
of 50 60
!
There are 12 moons in a year and 24 hours in a day. Divide twelve by 2 and we have 6, divide
24 by 2 and we have 12. We have:!
!
In that days=1, moons-12, hours=24.!
The Geometrical Explanation of Seconds
We have suggested the second is important as well in terms of the phases of the moon and the
earth and that these determined the calendar system we use. We further suggested there is a
connection of this to the structure found in geometry, and this is what we want to explore
further, here. We have all of this can be compactly written as:!
!
Where n=3, 4, 5, ,6 ,…!
We could evaluate this for n equal to other integers, or even the numbers, but these produce
the special triangles, and geometries we are most interested. Thus we will begin by pointing
out that!
Equation 11. 3 X 4 X 5 X 6 = 360!
The amount of degrees into which we divide a circle and that, as such it approximates the
number of days in a year (1 revolution of the earth around the sun) and thus we see that!
moon s
2
hours ⋅ d ays
= 6
2cos(π /n) = 1, 2, Φ, 3, . . .
of 51 60
Equation 12. !
Represents days as well (The earth moves through about 1 degree a day in its orbit around the
sun) by solving it for n this and the following are what might be characteristic equations of
habitable star systems:!
Equation 13. !
Where,!
!
Which correspond respectively to:!
!
Which are the unit triangle, unit the square, the unit regular pentagon, and the unit regular
hexagon.!
The Arrival of Mammalian Life
The asteroid belt seems to have been the source of an asteroid that hit Earth in the Yucatan 65
million years ago leading to the extinction of the dinosaurs and giving mammalian life a chance
to evolve into intelligence. Is this part of a natural process that takes place in other star
systems?!
The asteroid belt is about 1AU thick at 2.2 to 3.2 AU from the sun, the earth-sun separation
being 1 AU. Carbon-12 the basis of life as we know it is 12.01 g/mol. It is made in stars from
Beryllium-8 is 8.0053051 g/mol. Thus we have!
Equation 14. !
Equation 15. !
Delta x equal to 1 AU is both Earth and The Asteroids. Mars is at 1.52AU. Delta X cancels with
the Earth leaving Mars equal to carbon to beryllium, which is life, Does this say we Mars was
made available to Humans to colonize?!
!
Or, does it mean we need to put bases on the moon to mine Helium-3 as a clean, renewable
energy source.!
Application of the Theory
2cos(π /n) = 1, 2, Φ, 3, . . .
days = cos
−1
(y/2)
y = 1, 2,
5 + 1
2
, 3, . . .
n = 3,4,5,6,...
12
C
8
Be
=
Mars
Ear th
Δx
Δx = 1AU
12
C =
4
He +
8
Be
of 52 60
Kepler-442b is a larger than earth rocky planet orbiting a K-type main sequence star with a
mass of 0.61 solar masses, luminosity 12% that of the sun and 0.60 solar radius. It is about 2.9
billion years old compared to thus Sun at 4.6 billion years. It is at a distance of about 1,206
light years in the constellation Lyra. The planet orbits at 0.409 AU with an orbital period of
112.3 days (112 days 9 hours 10 minutes 24 seconds). It is 2.3 earth masses and 1.34 earth
radius. It receives 70% the light the earth receives from the sun.!
As it would turn-out this exoplanet is tidally locked with the star it orbits being closer in to its
star than the earth making its day perhaps weeks to months long. In so far as we don’t know
the length of its day we cannot determine the orbital period of any moon it might have in terms
of such a day. Nor do we even know if it has a moon to define a month from which we can infer
the number of months in a year on the planet. All we really know is it has a year 112.3 earth
days.!
However, let us speculate upon a scenario from what we have developed so far. Equation 9
was derived with the condition:!
Equation 16. !
We might formulate another scenario, While gold (Au) and silver (Ag) are metal used for making
religious and ceremonial jewelry so are silver (Ag) and copper (Cu). The thing all of these have
in common is that they are the finest malleable and ductile metals used fro the purpose of
making electrical wire as they are the finest conductors. Copper is among one of the first
metals used for the purpose of adornments and tools because it is soft enough to pound-out
without heating it, which was the method first discovered by our ancient ancestors. It was used
before smelting (separating from its ores) in its naturally occurring form. We thus have:!
Equation 17. !
Putting in the radius of the star as 0.6 the radius of the sun which is 696,000km and the molar
mass of silver is 107.87 and copper is 63.55 we have as the orbital
distance for the moon of Kepler-44b. Thus with the astronomical unit of the the moon (LAU)
being one corresponds to a period of 29.3 days. We have!
!
!
!
The period of the moon would be almost exactly 15 days. This connects the moon of this
planet to the moon of our planet in that the moon goes through about 15 degrees in an hour
(360/24=15). This gives credence to something I had considered earlier…!
R
⊙
r
m
=
Au
Ag
R
⊙
r
m
=
Ag
Cu
r
m
= 246,023.3k m
R
⊙
r
m
= 1.6974
T =
(
246,023.3
384,399
)
3
= 0.512
T = (0.512)(29.53) = 15.1209days
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Ethnomusicology is a sub-discipline in archaeology. It studies the history of humans through its
musical instruments and music theory. When we discussed the 12 month year based on twelve
moons per earth orbital period, we noticed it coincided with elegant mathematics because 12
was evenly divisible by 1, 2, 3, 4, 6 making it abundant. 12 in flamenco is a form of six. 6/8 is a
form common to may cultures and a favorite for being lively and dynamic throughout India, the
Middle East, and Africa. But perhaps the most beloved in India is Tin Tal, a cycle of 16 which is
four, four times. Four is often play four times throughout the Earth’s cultures. Thus, if a planet
orbited a star that had sixteen moons per its year, the inhabitants might divide the circle into
400 degrees and have four four month seasons rather than four three month seasons. This
results in the following table:"
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!
The luminosity of the sun is:!
!
The separation between the earth and the sun is:!
!
The solar luminosity at the earth is reduced by the inverse square law, so the solar constant is:!
L
0
= 3.9 × 10
26
J/s
1.5 × 10
11
m
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!
That is the effective energy hitting the earth per second per square meter. This radiation is
equal to the temperature, to the fourth power by the steffan-bolzmann constant, sigma ( ),
can be called the temperature entering, the temperature entering the earth.!
intercepts the earth disc, , and distributes itself over the entire earth surface, , while
30% is reflected back into space due to the earth’s albedo, a, which is equal to 0.3, so!
!
!
But, just as the same amount of radiation that enters the system, leaves it, to have radiative
equilibrium, the atmosphere radiates back to the surface so that the radiation from the
atmosphere, plus the radiation entering the earth, is the radiation at the surface of
the earth, . However, !
!
And we have…!
!
!
!
!
!
a=0.3!
!
!
!
S
0
=
39 × 10
2
4π(1.5 × 10
11
)
= 1,370wat ts/m eter
2
T
e
σ
T
e
S
0
πr
2
4πr
2
σ T
e
4
=
S
0
4
(1 − a)
(1 − a)S
0
(
πr
2
4πr
2
)
σ T
a
4
σ T
e
4
σ T
s
4
σ T
a
4
= σ T
e
4
σ T
s
4
= σ T
a
4
+ σ T
e
4
= 2σ T
e
4
T
s
= 2
1
4
T
e
σ T
e
4
=
S
0
4
(1 − a)
σ = 5.67 × 10
−8
S
0
= 1,370
1,370
4
(0.7) = 239.75
T
e
4
=
239.75
5.67 × 10
−8
= 4.228 × 10
9
T
e
= 255Kelvin
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So, for the temperature at the surface of the Earth:!
!
Let’s convert that to degrees centigrade:!
Degrees Centigrade = 303 - 273 = 30 degrees centigrade!
And, let’s convert that to Fahrenheit:!
Degrees Fahrenheit = 30(9/5)+32=86 Degrees Fahrenheit!
In reality this is warmer than the average annual temperature at the surface of the earth, but in
this model, we only considered radiative heat transfer and not convective heat transfer. In
other words, there is cooling due to vaporization of water (the formation of clouds) and due to
the condensation of water vapor into rain droplets (precipitation or the formation of rain).!
The program in C is as follows:!
#include <stdio.h>
#include<math.h>
int main(int argc, const char * argv[]) {
{
float s, a, l, b, r, AU, N, root, number, answer, C, F;
printf("We determine the surface temperature of a planet.\n");
printf("What is the luminosity of the star in solar luminosities?
");
scanf("%f", &s);
printf("What is the albedo of the planet (0-1)?" );
scanf("%f", &a);
printf("What is the distance from the star in AU? ");
scanf("%f", &AU);
r=1.5E11*AU;
l=3.9E26*s;
b=l/(4*3.141*r*r);
N=(1-a)*b/(4*(5.67E-8));
root=sqrt(N);
number=sqrt(root);
answer=1.189*(number);
printf("\n");
printf("\n");
printf("The surface temperature of the planet is: %f K\n",
answer);
C=answer-273;
F=(C*1.8)+32;
printf("That is %f C, or %f F", C, F);
T
s
= 2
1
4
T
e
= 1.189(255) = 303Kelvin
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printf("\n");
float joules;
joules=(3.9E26*s);
printf("The luminosity of the star in joules per second is:
%.2fE25\n", joules/1E25);
float HZ;
HZ=sqrt(joules/3.9E26);
printf("The habitable zone of the star in AU is: %f\n", HZ);
printf("Flux at planet is %.2f times that at earth.\n", b/1370);
printf("That is %.2f Watts per square meter\n", b);
printf("\n");
printf("\n");
}
return 0;
}
Let us run our program for Kepler-442b!
Here we use a single atmospheric layer with no
convection for the planet to be in an equilibrium
state. That is to say, the temperature stays
steady by heat gain and loss with radiative
heat transfer alone.
The habitable zone is calculated using the idea
that the earth is in the habitable zone for a
star like the Sun. That is, if a star is 100
times brighter than the Sun, then the habitable
zone for that star is ten times further from
it than the Earth is from the Sun because ten
squared is 100.
We determine the surface temperature of a planet.
What is the luminosity of the star in solar luminosities? 0.12
What is the albedo of the planet (0-1)?0.3
What is the distance from the star in AU? 0.409
The surface temperature of the planet is: 279.523163 K
That is 6.523163 C, or 43.741692 F
The luminosity of the star in joules per second is: 4.68E25
The habitable zone of the star in AU is: 0.346410
Flux at planet is 0.72 times that at earth.
That is 989.67 Watts per square meter
Ideally if the nearest star to us was similar to our Sun it would be best to start there. And
indeed the nearest star system is composed of a three stars and two of them are comparable
to the sun, in fact almost identical. But being a triple system, with which one of the stars has
been found to have a planet in its habitable zone, it can only be solved with equations other
than ours which work only for a single star with planets small in comparison to to its mass. The
triple star system I speak of is Alpha Centauri A, B, and C.!
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Alpha Centauri is the nearest of the stars to us and is a triple system designated Alpha
Centauri A, B, And C. Alpha Centauri A and B are much like the sun, similar in mass and
luminosity. Alpha Centauri A is is 1.1 solar masses and Alpha Centauri B is 0.907 solar masses.
They orbit a common center, which can be taken as the focus of an ellipse at the center of
gravity between the two. The period is 79.91 years. The eccentricity of the orbit is high and is
given as e=0.519.!
Alpha Centauri A is thought to have a Neptune sized planet in its habitable zone. While A and B
orbit one another their closest approach is 11.2 AU, and furtherest is 35.6 AU, but C (Proxima
Centauri) has its orbit very far from these two at 0.21 light years, a whopping 13,000 AU. The
11.2 and 35.6 AU for A and B are distances like Saturn and Pluto from the Sun, respectively.!
With the launching of the James Webb Space Telescope a lot of new data should be coming in
for planets and their moons, and we will have more to work with.!
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Conclusion 3
(Sexagesimal)
(Base 10 counting system, decimal system)
( )
If two lines each meet with a base at 60 degrees then they meet to make an apex that forms
another 60 angle. The is the equilateral triangle. Only the triangle is a self stabilizing pattern
because each side stabilizes the opposing two sides. (Buckminster Fuller, Synergetics).
But ultimately dividing things by 60 ultimately comes from the the Babylonians who inherited it
from the Sumerian counting system which was sexagesimal (base 60) perhaps because of its
convenience for dividing fractions because it is the smallest number divisible by 1, 2, 3. 4, 5, 6,
12, 10, 15, 20, 30 evenly."
60 × 60 = 3600
3600
10
= 360
3 × 4 × 5 × 6 = 360
2cos(π /n) = 1, 2, Φ, 3, . . .
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The Author!